9.4 The trapezium rule and limits on the estimate — Practise

Work through these on your own. Write down the strip width and every height before you substitute into the rule.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Use the trapezium rule with four strips to estimate $\displaystyle\int_0^2 x^2\,dx$ .
Find the exact value of $\displaystyle\int_0^2 x^2\,dx$ .
State whether the trapezium estimate $2.75$ for $\displaystyle\int_0^2 x^2\,dx$ is an overestimate or an underestimate, and give a reason.
Use the trapezium rule with eight strips to estimate $\displaystyle\int_0^2 x^2\,dx$ .
Using four strips and rectangles, give a lower limit and an upper limit for $\displaystyle\int_0^2 x^2\,dx$ .
Use the trapezium rule with four strips to estimate $\displaystyle\int_0^1 \frac{1}{1 + x}\,dx$ .
Find the exact value of $\displaystyle\int_0^1 \frac{1}{1 + x}\,dx$ .
State whether the four-strip estimate $0.6970$ for $\displaystyle\int_0^1 \frac{1}{1 + x}\,dx$ is too large or too small, and give a reason.
Using four strips and rectangles, give a lower limit and an upper limit for $\displaystyle\int_0^1 \frac{1}{1 + x}\,dx$ .
A car's speed is recorded every second: $0, 3, 5, 6, 6$ metres per second at $t = 0, 1, 2, 3, 4$ seconds. Estimate the distance travelled.
For the speed table in the previous question, give a lower limit and an upper limit for the distance travelled.
For $f(x) = x^2$ , find $f''(x)$ and use its sign to explain why the trapezium rule overestimates $\displaystyle\int_0^2 x^2\,dx$ .
A pond is sounded at $2$ metre intervals across a straight line: the depths are $0, 1.4, 1.9, 1.2, 0$ metres. Estimate the cross-sectional area.
For the pond in the previous question, give a lower limit and an upper limit for the area.

Answers and concise solutions

1. The estimate

Answer: Here $h = 0.5$ and the heights are $0, 0.25, 1, 2.25, 4$ , so the estimate is $\dfrac{0.5}{2}\bigl[0 + 4 + 2(0.25 + 1 + 2.25)\bigr] = 2.75$ .

2. The exact value

Answer: $\left[\dfrac{x^3}{3}\right]_0^2 = \dfrac{8}{3} = 2.6667$ to four decimal places.

3. The comparison

Answer: $2.75 > 2.6667$ , so the estimate is too large: an overestimate. The second derivative is $2 > 0$ , so the curve bends upwards and every chord lies above the curve.

4. The estimate

Answer: Here $h = 0.25$ , and the interior heights add to $8.75$ . The estimate is $\dfrac{0.25}{2}\bigl[4 + 17.5\bigr] = 2.6875$ . It is closer to $\dfrac{8}{3}$ than the four-strip estimate.

5. The limits

Answer: The left-hand heights give the lower sum $\dfrac{1}{2}(0 + 0.25 + 1 + 2.25) = 1.75$ . The right-hand heights give the upper sum $\dfrac{1}{2}(0.25 + 1 + 2.25 + 4) = 3.75$ . So $1.75 < \dfrac{8}{3} < 3.75$ .

6. The estimate

Answer: Here $h = 0.25$ and the heights are $1, 0.8, 0.6667, 0.5714, 0.5$ . The estimate is $\dfrac{0.25}{2}\bigl[1.5 + 2(2.0381)\bigr] = 0.6970$ to four decimal places.

7. The exact value

Answer: $\Bigl[\ln(1 + x)\Bigr]_0^1 = \ln 2 - \ln 1 = \ln 2 = 0.6931$ to four decimal places.

8. The comparison

Answer: The estimate is larger than $\ln 2 = 0.6931$ , so it is an overestimate. Also $\dfrac{d^2y}{dx^2} = \dfrac{2}{(1 + x)^3} > 0$ on $[0, 1]$ , so the curve bends upwards and each chord lies above the curve.

9. The limits

Answer: The function decreases, so the lower sum uses the right-hand heights: $\dfrac{1}{4}\left(\dfrac{4}{5} + \dfrac{2}{3} + \dfrac{4}{7} + \dfrac{1}{2}\right) = 0.6345$ . The upper sum uses the left-hand heights: $\dfrac{1}{4}\left(1 + \dfrac{4}{5} + \dfrac{2}{3} + \dfrac{4}{7}\right) = 0.7595$ . So $0.6345 < \ln 2 < 0.7595$ .

10. The distance

Answer: The distance is the area under the speed–time graph. With $h = 1$ , $s \approx \dfrac{1}{2}\bigl[0 + 6 + 2(3 + 5 + 6)\bigr] = 17$ metres.

11. The limits

Answer: The speed increases, so the lower sum uses the left-hand heights: $1(0 + 3 + 5 + 6) = 14$ metres. The upper sum uses the right-hand heights: $1(3 + 5 + 6 + 6) = 20$ metres. So the distance lies between $14$ and $20$ metres, and the estimate $17$ is the midpoint.

12. The second derivative

Answer: $f'(x) = 2x$ , so $f''(x) = 2 > 0$ on the whole interval. A curve with $f''(x) > 0$ bends upwards, so each chord lies above the curve and the trapezium estimate is too large.

13. The area

Answer: The readings are equally spaced with $h = 2$ , so $A \approx \dfrac{2}{2}\bigl[0 + 0 + 2(1.4 + 1.9 + 1.2)\bigr] = 9$ square metres.

14. The limits

Answer: The smaller reading on each strip gives the lower sum $2(0 + 1.4 + 1.2 + 0) = 5.2$ square metres. The larger reading on each strip gives the upper sum $2(1.4 + 1.9 + 1.9 + 1.2) = 12.8$ square metres. So the area lies between $5.2$ and $12.8$ square metres.